[proofplan] We use the continuous functional calculus for the normal element $a$: it identifies $C(\sigma_A(a))$ with the unital commutative $C^*$-subalgebra $C^*(a,1_A)\subset A$ by a unital isometric $*$-isomorphism. The multiplicative, adjoint, and norm identities are exactly the multiplicativity, $*$-preservation, and isometry of this isomorphism. For the spectral mapping formula, we compute the spectrum of the function $f$ inside $C(\sigma_A(a))$ directly as its range, transfer that spectrum through the isomorphism, and then use spectral permanence to regard the result as the spectrum in the ambient algebra $A$. [/proofplan]

[step:Introduce the functional calculus isomorphism] Let $B:=C^*(a,1_A)$ denote the unital $C^*$-subalgebra of $A$ generated by $a$ and $1_A$. Since $A$ is a unital complex Banach algebra, $\sigma_A(a)$ is a nonempty compact subset of $\mathbb C$, so $C(\sigma_A(a))$ is a unital commutative complex $C^*$-algebra under pointwise operations, pointwise conjugation, and the supremum norm. Since $a$ is normal, the continuous functional calculus theorem for normal elements applies. Thus there is a unital isometric $*$-isomorphism \begin{align*} \Phi_a:C(\sigma_A(a))\to B \end{align*} such that, for every polynomial $p$ in one complex variable and its conjugate variable, $\Phi_a(p)=p(a,a^*)$. By definition of the notation used in the theorem, for every $h\in C(\sigma_A(a))$ we write \begin{align*} h(a):=\Phi_a(h). \end{align*} [/step]

[step:Read multiplicativity and adjoints from the $*$-isomorphism]The product $fg\in C(\sigma_A(a))$ is the pointwise product defined by $(fg)(\lambda)=f(\lambda)g(\lambda)$ for $\lambda\in\sigma_A(a)$. Since $\Phi_a$ is multiplicative, \begin{align*} (fg)(a)=\Phi_a(fg)=\Phi_a(f)\Phi_a(g)=f(a)g(a). \end{align*} The function $\bar f\in C(\sigma_A(a))$ is the pointwise complex conjugate of $f$. The involution in $C(\sigma_A(a))$ is pointwise conjugation, so $f^*=\bar f$ in the $C^*$-algebra $C(\sigma_A(a))$. Since $\Phi_a$ is $*$-preserving, \begin{align*} (\bar f)(a)=\Phi_a(\bar f)=\Phi_a(f^*)=\Phi_a(f)^*=f(a)^*. \end{align*}[/step]

[guided]The continuous functional calculus is designed so that algebraic operations on continuous functions become the corresponding algebraic operations on the element $a$. We make this precise through the map \begin{align*} \Phi_a:C(\sigma_A(a))\to C^*(a,1_A). \end{align*} This map is a unital $*$-isomorphism, so it preserves multiplication and the involution. First consider multiplication. The function $fg:\sigma_A(a)\to\mathbb C$ is defined pointwise by \begin{align*} (fg)(\lambda)=f(\lambda)g(\lambda) \end{align*} for every $\lambda\in\sigma_A(a)$. Multiplicativity of $\Phi_a$ gives \begin{align*} (fg)(a)=\Phi_a(fg)=\Phi_a(f)\Phi_a(g)=f(a)g(a). \end{align*} The last equality is only notation: by definition, $\Phi_a(f)=f(a)$ and $\Phi_a(g)=g(a)$. Now consider the adjoint. In the commutative $C^*$-algebra $C(\sigma_A(a))$, the involution is pointwise complex conjugation. Therefore the adjoint of $f$ as an element of $C(\sigma_A(a))$ is the function $\bar f:\sigma_A(a)\to\mathbb C$ defined by \begin{align*} \bar f(\lambda)=\overline{f(\lambda)}. \end{align*} Because $\Phi_a$ is $*$-preserving, applying $\Phi_a$ after taking the adjoint is the same as taking the adjoint after applying $\Phi_a$. Hence \begin{align*} (\bar f)(a)=\Phi_a(\bar f)=\Phi_a(f^*)=\Phi_a(f)^*=f(a)^*. \end{align*}[/guided]

[step:Read the norm identity from isometry] The norm on $C(\sigma_A(a))$ is the supremum norm \begin{align*} \|f\|_\infty=\sup_{\lambda\in\sigma_A(a)}|f(\lambda)|. \end{align*} Since $\Phi_a$ is isometric and $f(a)=\Phi_a(f)$, \begin{align*} \|f(a)\|_A=\|\Phi_a(f)\|_A=\|f\|_\infty. \end{align*} [/step]

[step:Compute the spectrum of a continuous function as its range] Let $X:=\sigma_A(a)$, viewed as a compact subset of $\mathbb C$. Let $1_{C(X)}:X\to\mathbb C$ denote the constant function with value $1$. We claim that the spectrum of $f$ as an element of the unital $C^*$-algebra $C(X)$ is \begin{align*} \sigma_{C(X)}(f)=f(X). \end{align*} Indeed, fix $\lambda\in\mathbb C$. The element $f-\lambda 1_{C(X)}$ is invertible in $C(X)$ if and only if the [continuous function](/page/Continuous%20Function) \begin{align*} x\mapsto f(x)-\lambda \end{align*} has no zero on $X$. If it has no zero, then the reciprocal function \begin{align*} h:X\to\mathbb C,\qquad x\mapsto \frac{1}{f(x)-\lambda} \end{align*} is continuous and satisfies $h(f-\lambda 1_{C(X)})=1_{C(X)}$. Conversely, if $f(x_0)=\lambda$ for some $x_0\in X$, then every product $h(f-\lambda 1_{C(X)})$ vanishes at $x_0$, so $f-\lambda 1_{C(X)}$ cannot be invertible. Therefore $\lambda\notin\sigma_{C(X)}(f)$ exactly when $\lambda\notin f(X)$, which proves the claim. [/step]

[step:Transfer the spectrum back to the ambient algebra] Since $\Phi_a:C(\sigma_A(a))\to B$ is a unital algebra isomorphism, it preserves invertibility and therefore preserves spectra: \begin{align*} \sigma_B(\Phi_a(f))=\sigma_{C(\sigma_A(a))}(f). \end{align*} Using $\Phi_a(f)=f(a)$ and the preceding step, \begin{align*} \sigma_B(f(a))=f(\sigma_A(a)). \end{align*} Finally, $B=C^*(a,1_A)$ is a unital $C^*$-subalgebra of the unital $C^*$-algebra $A$ with the same unit, and $f(a)\in B$. By spectral permanence for unital $C^*$-subalgebras, \begin{align*} \sigma_A(f(a))=\sigma_B(f(a)). \end{align*} Combining the last two displayed equalities gives \begin{align*} \sigma_A(f(a))=f(\sigma_A(a)). \end{align*} This is the desired spectral mapping formula, and the proof is complete. [/step]